《ACM程序设计》书中题目Y

原创于 2017-03-21 20:35:51 发布 · 329 阅读

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本文介绍了一种基于描述前一元素生成后续元素的看数列算法实现。通过输入一系列数字，程序能够根据规则输出下一个序列。文章提供了一个C++实现示例，详细展示了如何逐个字符地遍历字符串并计算连续相同字符的数量。

Description
The look and say sequence is defined as follows. Start with any string of digits as the first element in the sequence. Each subsequent element is defined from the previous one by “verbally” describing the previous element. For example, the string 122344111 can be described as “one 1, two 2’s, one 3, two 4’s, three 1’s”. Therefore, the element that comes after 122344111 in the sequence is 1122132431. Similarly, the string 101 comes after 1111111111. Notice that it is generally not possible to uniquely identify the previous element of a particular element. For example, a string of 112213243 1’s also yields 1122132431 as the next element.

Input
The input consists of a number of cases. The first line gives the number of cases to follow. Each case consists of a line of up to 1000 digits.

Output
For each test case, print the string that follows the given string.

Sample Input

3
122344111
1111111111
12345

Sample Output

1122132431
101
1112131415

这个题还可以，就是查找，用string存数据再判断。

#include <bits/stdc++.h>
using namespace std;
int main()  
{  
    string s;    
    int len;  
    scanf("%d",&len);   
    for(int i=0;i<len;++i)  
    {  
        cin>>s;    
        char tmp = s[0];  
        int count = 1;  
        int idx = 0;   
        for(int j=1;j<s.size();++j)  
        {  
            if(s[j] == tmp)  
            {  
                ++count;  
            }  
            else  
            {  
                printf("%d%c",count,tmp);                    
                tmp = s[j];  
                count = 1;  
            }  
        }
        printf("%d%c\n",count,tmp);   
        //cout<<count<<tmp<<endl;  
    }    
    return 0;  
}